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Question

Find the number of $$5$$-digit odd numbers that can be formed using the integers from $$3$$ to $$9$$ if no digit is to occur more than once in any number


A
1440
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B
180
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C
360
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D
720
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Solution

The correct option is A $$1440$$
The digits from which we can choose digits is $$ 3,4,5,6,7,8,9 $$, that is a total of $$ 7 $$ digits

Since each desired number has to be odd , so we must have any of  $$ 3,5,7,9 $$ at the units place. This can be done in $$ 4 $$ ways
Then  ten thousands, thousands, hundreds, and tens  place can be filled up by remaining $$ 6$$ digits in $$ ^6P_4 = \dfrac {6! }{ (6-4)! } =\quad \dfrac { 6! }{ 2! } =\quad 6\times 5 \times 4 \times 3 =\quad 360 $$ ways
So, total number of $$ 5 $$  digit odd numbers that can be formed $$ = 360 \times 4 = 1440  $$

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